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Proof step 5 from proof absc-drop (previous step)

tree[parent(p)].dir?, parent(p) ∈ tree, q ⊑ p ⊦ p = ε, ∀ D1, dirs. ((abs|(tree[p, tree0], ino0, q, ino1) * D1))(dirs) ↔ ((abs|(tree, ino0, q, ino1) * D1))(dirs)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
tree[parent(p)].dir?, parent(p) ∈ tree, q ⊑ p ⊦ ((abs|(tree[p, tree0], ino0, q, ino1) * D))(dirs) ↔ ((abs|(tree, ino0, q, ino1) * D))(dirs), p = ε

Proof step 5 from proof absc-drop (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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